# Help with taking the derivative of any of these inverse tangent functions?

• March 29th 2009, 01:00 PM
dillonmhudson
Help with taking the derivative of any of these inverse tangent functions?
PS I KNOW THE FORMULAS I'M JUST NEW TO THIS AND STILL DON'T QUITE UNDERSTAND IT ALL (Headbang)

1. g(x) = 3arccos(x/2)

I got:
-3/[2*sqrt(4-x^2)]

As above without the '2'

2. f(x) = arctan (x/a)

I got:
[1/a] / [1 + x^2/a^2]

a / [a^2+x^2]

3. g(x) = arcsin (3x) / x
NO IDEA
[3x - sqrt(1-9x^2)*arcsin(3x)] / [x^2*sqrt(1-9x^2)]

4. h(t) = sin[arccos(t)]

-t/sqrt(1-t^2)

5. y = x*arccos(x) - sqrt(1-x^2)

arccos(x)

6. y = (1/2) * [ (1/2)*ln{x+1/X-1} + arctan(x) ]

1/[1-x^4]
• March 29th 2009, 05:34 PM
HallsofIvy
Quote:

Originally Posted by dillonmhudson
PS I KNOW THE FORMULAS I'M JUST NEW TO THIS AND STILL DON'T QUITE UNDERSTAND IT ALL (Headbang)

1. g(x) = 3arccos(x/2)

I got:
-3/[2*sqrt(4-x^2)]

As above without the '2'

The derivative of arccos(x) is $-1/\sqrt{1- x^2}$ so the derivative of g is $\frac{-3}{\sqrt{1-(x/2)^2}}(\frac{1}{2})= \frac{-3}{2\sqrt{1-(x/2)^2}}$. Now take that "2" inside the squareroot to get $\frac{-3}{\sqrt{4- 4(x/2)^2}}= \frac{-3}{\sqrt{4- x^2}}$.

Quote:

2. f(x) = arctan (x/a)

I got:
[1/a] / [1 + x^2/a^2]

a / [a^2+x^2]
Multiply both numerator and denominator of your answer by $a^2$.

Quote:

3. g(x) = arcsin (3x) / x
NO IDEA
[3x - sqrt(1-9x^2)*arcsin(3x)] / [x^2*sqrt(1-9x^2)]
Use the quotient rule: $(arcsin(3x)/x)'= [x(arcsin(3x))'- arcsin(3x)]/x^2$ $= \frac{\frac{3x}{\sqrt{1-9x^2}}- arcsin(3x)}{x^2}$.

Quote:

4. h(t) = sin[arccos(t)]

-t/sqrt(1-t^2)
The best way to do this is to note that is y= arccos(t), which is the same as t= cos(y), then we can imagine a right triangle with angle y, hypotenuse of length 1 and "opposite side" t. By the Pythagorean theorem, the "near side" has length $\sqrt{1- t^2}$ and $sin(y)= sin(arccos(t))= \sqrt{1- t^2}/1$. So $h(t)= \sqrt{1- t^2}= (1- t^2)^{1/2}$. Use the chain rule to differentiate that: $[(1-t^2)^{1/2}= (1/2)(1-t^2)^{-1/2}(-2t)= \frac{-t}{\sqrt{1- t^2}}$

Quote:

5. y = x*arccos(x) - sqrt(1-x^2)

arccos(x)
Use the product rule for x arccos(x): (x)' arccos(x)+ x(arccos(x))'= arccos(x)- $\frac{x}{\sqrt{1-x^2}}$ while the derivative of $-\sqrt{1- x^2}$ we just did above- its derivative is $\frac{x}{\sqrt{1- x^2}}$. You see that cancels the previous term, leaving only arccos(x).

Quote:

6. y = (1/2) * [ (1/2)*ln{x+1/X-1} + arctan(x) ]

The first thing I would do is write ln((x+1)/(x-1)) as ln(x+1)- ln(x-1) so its derivative is $\frac{1}{x+1}- \frac{1}{x-1}$. Of course the derivative of arctan(x) is $\frac{1}{1+ x^2}$.
y'= $\frac{1}{2(x+1)}- \frac{1}{2(x-1)}+ \frac{1}{1+x^2}$. (I've dropped the 1/2 multiplying the whole thing so I won't have to carry it along.) Add those fractions by getting the common denominator $2(x+1)(x-1)(x^2+1)= 2(x^2-1)(x^2+1)= 2(x^4- 1)$
$\frac{(x-1)(x^2+1)}{2(x^4-1)}- \frac{(x+1)(x^2+1)}{2(x^4-1)}+ \frac{2(x-1)(x+1)}{2(x^4-1)}$ $= \frac{x^3- x^2+ x- 1- x^3- x^2- x- 1+ 2x^2- 2}{2(x^4-1)}$ $= \frac{-2}{x^4-1}$.
Putting back the 1/2 I dropped gives $\frac{-1}{x^4-1}= \frac{1}{1- x^4}$.